Radial solutions of scaling invariant nonlinear elliptic equations with mixed reaction terms

We study global properties of positive radial solutions of --$\Delta$u = u p + M ||u| 2p p+1 in R N where p>1 and M is a real number. We prove the existence or the non-existence of ground states and of solutions with singularity at 0.


Introduction
The aim of this article is to study local and global properties of positive radial solutions of the equation − ∆u = |u| p−1 u + M |∇u| 2p p+1 , (1.1) in R N or R N \ {0} where p > 1 and M is a real parameter. This is a particular case of the following class of equations − ∆u = |u| p−1 u + M |∇u| q , (1.2) where q > 1 which has been the subject or many works in the radial case when M < 0, where a basic observation is that the two terms |u| p−1 u and M |∇u| q are in competition. The first work in that case is due to Chipot and Weissler [14] who, in particular, solved completely the case N = 1, then Serrin and Zou [19] performed a very detailed analysis. Much less is known in the case M > 0. Under the scaling transformation T k defined for k > 0 by Therefore, if q = 2p p+1 , (1.2) can be reduced to − ∆u = |u| p−1 u ± |∇u| and if q > 2p p+1 , the limit equation of (1.8) when k → 0 is the Riccati equation the exponent q is dominant. In [14] and [19] most of the study deals with the case q = 2p p+1 . In the critical case i.e. when q = 2p p + 1 , (1. 10) then not only the sign of M but also its value plays a fundamental role, with a delicate interaction with the exponent p. Notice that an equivalent form of (1.1) is with λ > 0. In the critical case first studies in the case M < 0 are due to Chipot and Weissler [14] for N = 1. The case N ≥ 2 was left open by Serrin and Zou [19] and the first partial results are due to Fila and Quittner [16] and Voirol [22,23]. The case M > 0 was not considered.
The equation (1.1) is the stationary part of the associated parabolic equation ( 1.12) which is studied in [14] and [15], one of the aim being to find conditions for the blow-up of positive solutions. A general survey with several open problems can be found in [20].
In the non radial case an important contribution dealing with a priori estimates of local positive solutions of (1.2) and existence or non-existence of entire positive solution in R N is due to the authors [7]. In this paper we complete the results of [7] in presenting a quite exhaustive study of the radial solutions of (1.1) for any real number M .
The radial solutions of (1.1) are functions r → u(r) defined in (0, ∞) where they satisfy Because of the invariance of (1.13) under the transformation T k there exists an autonomous variant of (1.1) obtained by putting then (1. 16) Putting y(t) = −r p+1 p−1 u r (r), then (x(t), y(t)) satisfies the system (1. 17) We are mainly interested in the trajectories of the system which remain in the first quarter Q={(x, y) ∈ R 2 : x > 0, y > 0}. Among these trajectories, the ones corresponding to ground states, i.e. positive C 2 solutions u of (1.13) are defined on [0, ∞). They verify u r (0) = 0 and actually they are C ∞ on (0, ∞)). Using the invariance of the equation under T k all the ground states can be derived by scaling from a unique one which satisfies u(0) = 1. Since it is easy to prove that such a solution u is decreasing, in the variables (x, y), a ground state is a trajectory of (1.17) in Q, defined on R and satisfying lim  for all r > 0, (1.18) where X is a positive root of This equation plays a fundamental role in the description of the set of solutions of (1.13). The following constant, defined for N = 1, 2 and p > 1 or N ≥ 3 and 1 < p ≤ N N −2 has an important role in the description of the set roots of (1. 19), . (1.20) This set is described in the following proposition. N −2 . If p = N +2 N −2 these ground states are explicit and they satisfy lim r→∞ r N −2 u(r) = c > 0. There exist infinitely many singular solutions u ondulating around U M . Note that a ground state corresponds to a homoclinic orbit at 0 for system (1.17) and these singular solutions are cycles surrounding P M . We recall that an orbit of (1.17) which connects two different equilibria (resp. the same equilibrium) when t ∈ R is called heteroclinic (resp. homoclinic).
Without the radiality assumptions, and using a delicate combination of refined Bernstein techniques and Keller-0sserman estimate we have obtained in [ . (1.21) Then M is positive (resp. negative) if p > N +2 N −2 (resp. p < N +2 N −2 and we set µ = M ). It is easy to see that if M = M then the characteristic values of the linearized operator at P M are purely imaginary. Notice that M is positive (resp. negative) according p > N +2 N −2 (resp. p < N +2 N −2 ). Theorem A Let N ≥ 1, p > 1 and M > 0. -for M >M max there exists no radial ground state.
The values ofM min andM max appear as transition values for which the ground state still exists but it is smaller than the others at infinity; it is of order r 2−N instead of r − 2 p−1 . They are not explicit but they can be estimated in function of N and p. It is a numerical evidence that M min =M max in the phase plane analysis of system (1.17) and we conjecture that this is true. When M =M min orM max , the system (1.17) admits homoclinic trajectories. We prove that the system (1.17) admits a Hopf bifurcation when M = M . When p > N +2 N −2 we also prove the existence of different types of positive singular solutions In terms of the system (1.17) the 1) and 2-(i) correspond to the existence of a heteroclinic orbit in Q connecting P M to (0, 0) and (ii) to the existence of a cycle in Q surrounding P M .
When M is negative, the precise description of the trajectories of (1.17) depends also on the value of p with respect to N N −2 . It is proved in [7, Th. B, E] that for N ≥ 3 and 1 < p < N +2 The same conclusion holds if N ≥ 3, 1 < p ≤ N N −2 (or N = 2 and p > 1) and M > −µ * . We first consider the case p ≥ N N −2 for which there exists a unique explicit singular solution U M , and the results present some similarity with the ones of Theorem A.

2-If
such that (i) for M < −μ max there exist ground states u such that u(r) ∼ U M (r) when r → ∞.
(ii) for M = −μ min or for M = −μ max there exist ground states minimal at infinity in the sense that u(r) ∼ cr 2−N when r → ∞, c > 0.
(iii) for −μ min < M < 0 there exists no radial ground state.
Here also the value ofμ min ,μ max are not explicit and we conjecture that they coincide. The next result presents some similarity with Theorem A'.
there exists a unique (up to scaling) positive singular solution u of (1.13), such that u(r) ∼ U M (r) when r → 0 and u(r) ∼ cr 2−N when r → ∞ for some c > 0.
(ii) If −μ min < M < 0 there exist positive singular solutions u ondulating around U M on [0, ∞) and singular solution ondulating around U M in a neighbourhood of 0 and satisfying u(r) ∼ cr 2−N for some c > 0 when r → ∞.
In terms of the system (1.17), (i) corresponds to a heteroclinic orbit connecting P M and (0, 0), while (ii) to the existence of a periodic solution in Q around P M , and the existence of a solution in Q converging to (0, 0) at ∞ and having a limit cycle at t = −∞ which is a periodic orbit around P M .
The situation is more complicated when 1 < p < N N −2 and M < −µ * because there exist two explicit singular solutions U 1,M and U 2,M which coincide when M = −µ * .
Here againμ min andμ max appear as transition values for which the ground state still exists but it is smaller than the others at infinity: it behaves like U 1 instead of U 2 . The proof of this theorem is very elaborate in particular in the case N = 2. In the case N = 1 the result is already proved in [14]. The nonexistence of ground state, not necessarily radial for M > −µ * is proved in [1] and independently in [7] with a different method. In the radial case it was obtained much before in the case N = 1 in [14] and then by Fila and Quittner [16] who raised the question whether the condition −μ min < M < 0 is optimal for the non-existence of radial ground state. This question received a negative answer in the work of Voirol [22] who extended the domain of non-existence to −µ * − < M ≤ −µ * . The next result is the counterpart of Theorem C when dealing with singular solutions.
(ii) If M ≤ M < −µ * there exists a unique up to scaling positive singular solution u, such that u(r) = U 2,M (r) as r → 0 and u(r) = U 1,M (r) as r → ∞. Furthermore u(r) > U 1,M (r) for all r > 0.   In terms of the system (1.17) (i) corresponds to a heteroclinic orbit connecting P 1,M to (0, 0); (ii) to a heteroclinic orbit connecting P 2,M to P 1,M ; (iii) to a trajectory having a periodic orbit around P 2,M for limit cycle at −∞ and converging to P 2,M at ∞ and to a periodic orbit around P 2,M ; (iv) corresponds to homoclinic orbit at P 1,M ; (v) corresponds to a trajectory connecting (0, 0) at −∞ and either converging to P 2,M at ∞ or having a periodic orbit around P 2,M for limit set at ∞; (vi) corresponds to a heteroclinic orbit connecting from (0, 0) to P −µ * .
Acknowledgements This article has been prepared with the support of the collaboration programs ECOS C14E08 and FONDECYT grant 1160540 for the three authors.
2 General properties of the system 2.1 Reduction to autonomous equation and system

The standard reduction
We recall that if u is a C 3 function defined on some interval I ⊂ [0, ∞) verifying (1.13) and if then x satisfies the autonomous equation on ln(I) where K and L are defined in (1.16). Setting u r = −r − p+1 p−1 y(t), then (x(t), y(t)) satisfies where and we denote by H the vector field of R 2 with components H 1 and H 2 .    (1)) as y → ∞. If M < 0 and K < 0, ψ is still concave and increasing on ( K M p+1 p−1 , ∞) with the same asymptotic as above. We quote below the possible connected components of Q \ (L ∪ C).
These connected components are

Graphic representation of the vector field H
We present below some graphics of the vector field H associated to system (1.17). Figure 4: Figure 5: M < −µ * , K < 0.

Other reduction
The following change of unknowns, already used in [9] when M = 0, and valid if u r = 0, transforms (1.17) into a Kolmogorov system with vector field V = (V 1 , V 2 ) (2.6) Since σ and z are in factor the two axis {σ = 0} and {z = 0} are trajectories, actually not admissible for (2.2) in view of (2.5). The system is singular on these two axis however it can be desingularized in setting σ =σ 2k+1 and z =z 2k+1 for some integer k > p + 1, which transforms (2.6) into a new nonsingular Kolmogorov system, Therefore no other trajectory can intersect them in finite time and the quadrant Q := {(σ, z) : σ > 0, z > 0} is invariant. Furthermore σz = r 2 |u| p−1 . It is noticeable that if M = 0 the initial system is quadratic and regular.
The system (2.6) will be used in the most delicate cases. It corresponds to the differentiation of the initial equation (2.1).

Regular solutions and ground states
Definition 2.1 A regular solution of (1.13) is a C 2 solution defined on some maximal interval [0, r 0 ) satisfying u(0) = u 0 > 0 and u r (0) = 0. A ground state is a nonnegative C 2 solution defined on [0, ∞).
The existence and uniqueness of a regular solution is standard by the Cauchy-Lipschitz integral method. If u is a C 2 solution it satisfies u r < 0 on (0, r 0 ). Indeed r N −1 u r (r) is decreasing near 0, hence u r < 0 on some maximal interval (0, r 1 ) ⊂ (0, r 0 ) and u r (r 1 ) = 0 if r 1 < r 0 . If u(r 1 ) = 0 then u ≡ 0 by uniqueness. If u(r 1 ) > 0 then u rr < 0 near r 1 which would imply that u(r) < u(r 1 ) for r 1 − ≤ r < r 1 which contradict the negativity of u r on (0, r 1 ). Hence u(r 1 ) < 0 which implies that u r (r) < 0 on the maximal interval (0, r 2 ) where u > 0. Thus, if u is a ground state u r < 0 on (0, ∞). Hence the trajectory of a ground state expressed in the system (1.17) lies in Q and expressed in the system (2.6) it lies in the quadrant Q. It is easy to check that the regular solution u := u u 0 , such that u(0) = u 0 satisfies Under the scaling transformation T k , u can be transformed into the regular solution (1.13) u := u 1 satisfying u(0) = 1. If one considers the system (1.17) the transformation T k becomes the time shift which transforms t → (x(t), y(t)) into t → (x(t + ln k), y(t + ln k)), and the trajectory (x(t), y(t)) of (1.17) corresponding to a ground state is therefore uniquely determined and denoted by T reg and satisfies

Explicit singular solutions
Explicit self-similar solutions of (1.13), necessarily under the form u = Ar − 2 p−1 , play a fundamental role in the study, whenever they exist. The following result covers Proposition 1.
. Then the left-hand side of (2.12) is clear. Next we put we derive η ≤ 1 + a, which implies the right-hand side of (2.12).
For a sharper estimate, we have for M large enough, The estimates (2.13) follow.

Upper estimate of the regular solutions
We first recall the following estimate in the case M ≥ 0, consequence of the fact that the positive solutions of (1.1) are superharmonic and proved in a more general setting in [7, Prop. 2.1 ].
In particular there exists no ground state.
The next estimate is verified by any ground state, independently of the sign of M .
Proof. The trajectory T reg starts from (0, 0) and enters the region C. If it stays in C, then x is increasing on R. Since y < 2x p−1 , if x(t) tends to some finite limit as t → ∞, it implies that the limit set of the trajectory exists. It cannot be a cycle since x is monotone, thus it is one of the equilibrium of the system. Hence x(t) ≤ X j,M for j = 1 or 2 if K < 0 and M < −µ * or x(t) ≤ X M if K ≥ 0 and either M > 0 or −µ * ≤ M ≤ 0. This implies that (2.20) holds. If Therefore y t ≥ Cy p for some C > 0 which would imply that t → y 1−p (t)+C(p−1)t is increasing, which is impossible.
Next we suppose that the trajectory leaves C by crossing the line L. Since it cannot enters B through C (in the case K < 0, M ≤ −µ * ), it leaves C by intersecting L and we denote by t 1 the first time where T reg intersects L. Then x t (t 1 ) = 0 and x tt (t 1 ) = −y t (t 1 ) < 0. Therefore t 1 is a local maximum. Now the trajectory cannot cross again the half line (x, y) : In the same way, either y is increasing on R and since x t = 2 p−1 x−y and x is bounded, y cannot tend to infinity when t → ∞, thus y(t) → y 0 > 0, or y is not monotone and T reg crosses C at a first value t 2 , necessarily larger than t 1 and where x t (t 2 ) < 0. Then y tt (t 2 ) = p |x(t 2 )| p−1 x t (t 2 ) < 0 and t 2 is a local maximum of y. Therefore ∪ t≥t 2 (x(t), y(t)) remains in the subset of Q bordered by the portion of trajectory of T reg for t ≤ t 2 and {(x, y) : 0 < x ≤ x(t 2 ), y = y(t 2 )}. This implies that y(t) ≤ y(t 2 ) for all t ∈ R. Noticing that u(r) ≤ 1 since u is decreasing, we get the conclusion.
Remark. The above method does not give an explicit estimate of the upper bounds of x and y and such a bound can be estimated in some cases. If M ≥ 0 it follows from Proposition 2.3 that for any p > 1 there holds 2.4 Linearization of the system (1.17) near equilibria 2.4.1 Linearization at (0, 0) which admits the eigenvalues (a) Assume that N ≥ 3 and p > N N −2 , equivalently K > 0, Then (0, 0) is a saddle point. There exists a unique unstable trajectory T unst such that and more precisely, from (2.8), (2.26) From Definition 2.2 and the lines which follow, the unstable trajectory T unst coincides with the regular trajectory T reg . This is included in the region C for t < T 0 for some −∞ < T 0 ≤ ∞. There exists also a unique stable trajectory T st such that lim t→∞ (x(t), y(t)) = (0, 0) and lim , y(t)) = (0, 0), then T reg = T st and the corresponding solution is a ground state. The same conclusion holds, if T st ⊂ Q satisfies lim t→−∞ (x(t), y(t)) = (0, 0). Such a solution is called a homoclinic orbit at (0, 0). Because of the uniqueness of the stable and unstable trajectories of a saddle point, it is unique in the class of solutions satisfying (2.27). Equivalently the class of ground states u of (1.13) satisfying u(r) ∼ cr 2−N for some c > 0 is then a one parameter family characterized by u(0) = u 0 . (b) Assume that N ≥ 3 and 1 < p < N N −2 . Then K < 0 and 0 < λ 1 < λ 2 . Hence (0, 0) is a source and all the trajectories of (1.17) in some neighbourhood of (0, 0) converge to (0, 0) when t → −∞. Among those trajectories there exists one fast trajectory which satisfies It is actually the regular trajectory T reg . There exist also infinitely many slow trajectories which satisfy (c) If p = N N −2 , then K = 0 and λ 1 = 0 < λ 2 = N − 2. We still find the regular trajectory T reg associated to λ 2 and the corresponding eigenvector (1, 0). By the central manifold theorem corresponding to λ 1 there exists an invariant curve passing through (0, 0) with slope N −2. Using the matched asymptotic expansion method, one finds that if M < 0 there exists a solution x of (2.1) such that p−1 , with corresponding eigenspace (1, 0). The linearized problem is equivalent to equation q−1 for some real parameters a, b. Hence there exists infinitely trajectories of (2.1) tending to 0 when t → −∞ and they are tangent to (1, 0) at (0, 0). The regular trajectory T reg corresponds to b = 0 and the other trajectories correspond to singular solutions u of (1.1). They satisfy u(r) ∼ b ln r as r → 0 and there holds in the sense of distributions in B for some > 0.
Next we give a general result in case the system admits only one equilibrium in Q.
If u is a regular solution the following tetrachotomy occurs: (i) either lim r→∞ r N −2 u(r) = c for some c > 0, (ii) or u(r) ∼ U M (r) as r → ∞, (iii) or u(r) has an ω-limit cycle surrounding P M , (iv) or u(r) changes sign for some r > 0.
Proof. By assumption P M is the unique equilibrium. The trajectory T reg starts from (0, 0) and remains in the region C where x t , y t > 0 for t ≤ t 0 ≤ ∞. If t 0 = ∞, u is a ground state, hence it is bounded from Proposition 2.4. Its ω-limit set is non-empty. Because x and y are monotone, it converges when t → ∞ to some point which is necessarily P M . If t 0 < ∞, then at t = t 0 the trajectory leaves C through L since it cannot enter in B, and it enters the region D where x t < 0, y t > 0. Moreover x t (t 0 ) = 0 and x(t 0 ) > X M . Then three possibilities occur: (α) either x(t) → X M monotonicaly when t → ∞; thus the trajectory converges to P M . (β) either x(t) → 0 monotonicaly when t → ∞. Since (0, 0) is a saddle point, then T reg =T st . This implies that T reg is a homoclinic trajectory at (0, 0). (γ) or there exists t 1 > t 0 such that x t (t 1 ) = 0. Then x(t 1 ) < X M . Hence T reg enters the region B and by continuity there exists t < t 0 such that x(t ) = x(t 1 ) and y(t ) < y(t 1 ). Therefore the bounded region of R 2 bordered by the segment I = {(x, y) : x = x(t 1 ), y(t ) < y < y(t 1 )} and the portion of T reg defined by {(x(t), y(t)) ∈ T reg : t ≤ t ≤ t 1 } is positively invariant (notice that x t > 0 on I) and it contains P M and no other equilibrium. Therefore either the trajectory converges to P M or it admits an ω-limit cycle which is a closed orbit surrounding P M .

Linearization at a fixed point P
The eigenvalues of its matrix are the roots of Since Y M is a positive root of (2.15), we get hence M , well defined for N ≥ 2, is given by . (2.31) That is (1.21).
Remark. We see that M > 0 (resp. M < 0) if and only if N ≥ 3 and p > N +2 is a center for the system (1.17) associated to N −2 and that there exist infinitely many cycles turning around P 0 with equation It can be verified that E < 0, in particular using the function F defined in (2.35). For N ≥ 2 and 1 < p < N N −2 , there always holds M ≤ −µ * and more precisely But |L| − 2 |K| = N −2 p−1 and the conclusion follows. If N = 2 we just replace < by = in the above series of equivalences.
N −2 and M = M > 0 then P M is a weak sink and a Andronov-Hopf bifurcation point. If p < N +2 N −2 and M = M < 0 then P M is a weak source.
Proof. We recall that a weak sink is an asymptotically stable equilibrium which attracts the nearby points as t → ∞ at a rate slower than the usual exponential rate. A weak source is a weak sink of the system obtained by changing t into −t (see [17,Chap. 9]). We writex = x−X M , y = y − Y M and obtain the new nonlinear system (2.34) In order to compute the Lyapunov coefficients we transform the system by setting By integrating the first line and using the expansion of h, we obtain This can be written in the following way By [17,Th. 9.2.3], the Lyapunov coefficient is given by Λ = ν 2,0 + 3ν 0,3 + ν 1,1 (ν 2,0 + ν 0,2 ) which yields by computation If L > 0 (resp. L < 0) P M is a weak sink (resp. a weak source).

Energy and Lyapunov functionals for system
(2.36) Next we construct a Lyapunov functional adapting the method initiated by [2] and already used in [3] and [4].
Proof. We recall the ansatz introduced in [2] for finding a Lyapunov function for a system of the form In the case of system (1.17), h(x) = 2x p−1 , and we find L(x, y) = J (x, y). Then (2.38) and the conclusion follow.
Moreover H is starshapped with respect to 0 and we set In [7] we also used a function introduced in [19] for equation (1.2). When q = 2p p+1 it reduces to p+3 . Since r = e t , we find that in Q, The function satisfies the relation Note that U has a constant sign in Q if LM < 0. Proof. We use the expansion (2.26) and deduce, for Suppose that the trajectories intersect for a first time at some point (x 0 , y 0 ) below L. Since the system is autonomous, there will exist two solutions of the systems relative to M and M satisfying at the same time t    Proof. Step 1: Assume N N −2 < p ≤ N +2 N −2 . The linearized system is given in (2.28). Because M > 0, the product of the characteristic roots given by equation (2.29) is positive since it is given by

Behaviour near equilibrium
The sum (or the real part) of the characteristic roots is equal to Hence P M is a source. Step 3: If M = M , and p > N +2 N −2 , then P M is a weak sink by Lemma 2.7. The appearance of the limit cycle, which is the called the Andronov-Hopf bifurcation, occurs for M > M when M − M is small enough (see [17,Chap. 9]).
Remark. The product of the characteristic roots is also expressed by Hence it is positive for any M ≥ 0, p > 1.
Next we give some sufficient conditions for nonexistence of a periodic solution or a homoclinic orbit at (0, 0). Proof. If γ is a non-trivial closed orbit it corresponds either to a T -periodic solution or a solution such that lim The function V defined in (2.37) is monotone and it satisfies either Hence it is constant and by (2.38) it implies y(t) = 2 p−1 x(t) for all t, a contradiction. Next we suppose p > N +2 N −2 , 0 < M ≤ M and that there exists a T -periodic solution (x(t), y(t) with trajectory γ ⊂ Q surrounding P M , hence P M belongs to the bounded connected component Γ of R 2 \ {γ} bordered by γ. Since P M is a sink or a weak sink by Lemma 3.1, there exists a neighbourhood O of P M such that all the trajectories issued from O converge to P M as t → ∞. Hence any trajectory issued from O, necessarily contained in Γ, has an α-limit set in Γ which is either a stationary point different from P M , which is excluded, or a limit cycle {(x(t), y(t)} t∈[0,τ ) := γ ⊂ Γ (τ is its period). This limit cycle is not stable, hence, by Floquet's theory We perform the change of unknownsx = x − X M ,ȳ = y − Y M used in Lemma 2.7 which leads to the system (2.33). The explicit value of the remaining term is where Φ and Ψ are defined accordingly. It is positive by convexity because M > 0. Since from Using the equation (2.15) satisfied by Y M it yields Combining (3.1) and (3.3), we obtain which contradicts (3.4).
Remark. Up to changing the sense of variation of V(t), the proof of the first assertion shows that there exists no closed orbit in Q if M < 0 and p ≥ N +2 N −2 . However the proof of the second assertion is not valid when M < 0.
The nonexistence of any periodic solution can also be proved when the equilibrium is a node (i.e. the two characteristic values are real with the same sign).
Hence (3.5) is satisfied if one of the following conditions holds: Using (2.15) we find (3.8) Concerning the lower bound in (i), there holds Then we can define M 1 by which leads to Hence Next we prove 3) by adapting an argument introduced in [12] for quadratic systems. We return to system (2.33) that we write under the form which defines a, c, d with Φ and Ψ given by (3.13), and the trinomial (2.29) for characteristic values endows the form In the range of values of M , the discriminant D = (d + a) 2 − 4(ad + c) = (d − a) 2 − 4c is positive. We consider the intersection of a straight line passing through P M with equationȳ = Ax with a trajectory (x(t),ȳ(t)). Then We can choose A = 0 such that A 2 + (d − a)A + c = 0 since D > 0. Since Φ and Ψ achieves positive values, we derive from the expression of h that U > 0 forx = 0. This proves that any closed orbit around P M or passing by P M can intersect only one time which is a contradiction. Proof. Assume that T reg remains in Q, by Lemma 2.5 we have three possibilities: (α) either T reg converges to P M when t → ∞, which is impossible since P M is a source, (β) or T reg has a limit cycle at ∞, and this is impossible by Lemma 3.2, (γ) or T reg converges to (0, 0) when t → ∞, hence it is a homoclinic orbit. The function V defined in (2.37) is decreasing by Lemma 2.8. Since V(−∞) = V(∞) = 0, it is identically 0 and so is V t . This implies that 2x(t) p−1 − y(t) = 0 for all t which is a contradiction. Hence T reg does not remain in Q.

Existence or nonexistence of ground states
We denote by Ξ the connected region of Q bordered by the semi-axis {(x, y) : x = 0, y > 0} and T reg . Since H is outward on the semi-axis, Ξ is negatively invariant. By Section 3.4.1, which implies u(r) ∼ cr 2−N for some c > 0. Its α-limit set α(T st ) cannot be a limit cycle as we have seen it above. If it contains (0, 0) it implies again that V, which is monotone, is equal to 0, hence V t ≡ 0 and 2x(t) p−1 − y(t) ≡ 0, which is impossible. Hence α(T st ) contains P M . Since P M is a source it implies that T st converges to P M when t → −∞. Proof. If 0 < M < M (resp. M = M ), P M is a sink (resp. a weak sink). Suppose first that the trajectory T reg does not stay in Q, then it leaves Q at some point (0, y s ) with y s > 0. As a consequence, the stable trajectory T st at (0, 0) remains in the negatively invariant region Ξ defined in the proof of Proposition 3.4. Since it cannot converge to P M when t → −∞ it admits a limit cycle surrounding P M which contradicts Lemma 3.2. Therefore T reg ⊂ Q and, again using Lemma 3.2, either it converges to P M when t → ∞ and the proof is complete, or to (0, 0) and T reg = T st is a homoclinic trajectory. The trace of the linearized system (2.23) at (0, 0) is equal to 2 p−1 − K = −L < 0. Therefore, from [17, Th. 9.3.3] the connection is attracting and the trajectories inside the bounded region T bordered by the homoclinic trajectory T reg spiral towards it when t → ∞. Hence any such trajectory inside T either has a limit cycle when t → −∞ which is impossible by Lemma 3.2 or converges to P M which is also impossible. Consequently there exists no homoclinic trajectory at (0, 0) which ends the proof.
Next we study the case where M is large enough. We have already proved in [7] that for any p > 1, there exists M † = M † (N, p) > 0 (see introduction) such that if M > M † there exists no ground state, radial or non-radial. In the radial case we have a more precise result.  Theorems A and A' follow from the previous results.
Remark. It is a challenging question to prove that there is unique M such that there is a homoclinic trajectory at (0, 0). Up to now all we can prove is that if there exist two parameters 0 < M 1 < M 2 such that for each of them there exists a homoclinic trajectory of (0, 0) in Q T 4 Study of ground states of (1.17) when M < 0 We will distinguish the cases p ≥ N N −2 where system (1.17) admits a unique non-trivial equilibrium and 1 < p < N N −2 where the existence of zero, one or two equilibria depends on the value of M with respect to −µ * defined in (1.20). In order to avoid confusion we set . (4.1) Since K ≥ 0 the product of the roots is positive; the real part of the roots is positive if and only Notice that there always hold |L| + 2 − 2 √ N − 1 < |L| since N ≥ 3. The first condition in (4.4) requires |L| N −2 , the conclusion follows and M 0 and M 1 are given by (3.8) and (3.11).
The most intricate case corresponds to 1 < p < N N −2 or N = 1, 2 where there may exist 0, 1 or 2 equilibria. 2,M is increasing. We first consider the linearized operator at P 1,M . The product of the roots of (4.2) Hence P 1,M is a saddle point. and Hence P 2,M is a source. If M < M , then the sign in (4.5) is reversed and Thus P 2,M is a sink. By Lemma 2.7 P 2,M is a weak source, and assertion 2 is proved. Finally, if N = 2 and M < −µ * = M from Lemma 2.6, therefore P 2,M is a sink. Next we look for conditions which insure that P 2,M is a node. The characteristic roots are real if and only one of the two conditions (3.6) where Y M is replaced by Y 2,M holds: (4.8) , there holds Therefore P 2,M is always a sink and the discriminant of (2.29) is equal to 2p hence the characteristic roots are negative.
Notations. Under the assumptions of Proposition 4.2 there exist two stable trajectories T 1,j st , j=1,2, converging to P 1,M when t → ∞, associated to a negative characteristic value λ; the common slope at P 1,M is 2 p−1 + |λ|. We assume that T Next we look for the existence of limit cycles. Since M < 0, we cannot argue using the convexity argument used in Lemma 3.2. We use system (2.6) which also has a convexity property as we see it in the proof below. Proof. For 1, assume that there exists a periodic trajectory (x(t), y(t) surrounding P M . By Green's formula, In the phase plane (σ, z) the equilibrium P M becomes Similarly P 2,M becomes P 2,M = (σ 2,M , z 2,M ) = ( and P M and P 2,M are sources. Hence, any trajectory converging to this point P M (or P 2,M ) admits an omega-limit cycle γ ⊂ Γ 0 . Consequently this cycle is not unstable which implies that the Floquet integral relative to this T-periodic solution is nonpositive: The computation gives We set σ = σ M + σ and z = z M + z (the computation would be the same with (σ M , z M ) replaced by (σ 2,M , z 2,M )), then

By addition
Integrating on a period, we get Therefore Lemma 4.4 Let p > 1. If N ≥ 2 and M ≤ −µ * (1) or N = 1 and M < −µ * (1), then there exists a ground state u. Furthermore there holds As a consequence, the corresponding trajectory T reg = {(x reg (t), y reg (t)} t∈R does not converge to (0, 0) when t → ∞. If 1 < p < N N −2 and N ≥ 3 or p > 1 and N = 1, 2, T reg does not converge to P 1,M when t → ∞.
Next we give an alternative proof of a result of [16]. Proof. If −µ * < M < 0, the only equilibrium of (1.17) is (0, 0) and it is a source by Section 3.4.1. If there exists a ground state then the trajectory T reg remains in Q. It is bounded by Proposition 2.4 hence it admits an ω-limit set which either contains an equilibrium or is a periodic orbit. The two possibilities are excluded.
Next we study the case M ≤ −µ * . In particular we cover the case M = −µ * studied in [22] with a different proof, using the energy function Z defined in (2.41).
(i) Suppose that T reg converges to P 1,M with x(t) increasing, hence (x(t), y(t) is below L and y(t) is also increasing. Then from (2.43) r → e (ii) Suppose that T reg intersects L at a point (xM ,ỹM ) between (0, 0) and P 1,M , i.e.ỹM < Y 1,M . (a) If M < −µ * , then consider the stable trajectory T 1,1 st at P 1,M which is below L: T 1,1 st cannot converge to (0, 0) when t → −∞; indeed it would be a unstable trajectory at the source point (0, 0) and since T reg is the unique fast unstable trajectory at this point (see Section 2.4.1), it is below T 1,1 st near zero and the two curve would intersect. Therefore T 1,1 st leaves Q through the semi-axis {(x, 0) : x > 0} at some at some x = x(τ ) and (x(t), y(t)) ∈ Q for t > τ . Settinḡ r = e τ , there holds Z(r) =r a u p+1 (r) > 0 and again r → e  (ii) If M = −µ * then T reg intersects L at some point (x, y) with x > X −µ * and leaves Q; there is no ground state.
Proof. (i) Suppose M < −µ * . If T reg remains below L, then x t > 0, y t > 0 and T reg converges to P 1,M or P 2,M The first limit is excluded by Lemma 4.6 and the second by assumption. Hence T reg intersects L. This intersection cannot occur between (0, 0) and P 1,M and between P 1,M and P 2,M since the vector field H is inward in the region B on this segment, so it occurs at some point (x, y) with x > X 2,M .

Ground states and large solutions when M < 0
The following result extends [7, Theorem B'] to a larger class of parameters M in the radial case.  13) in (a, b) for a < b tending to infinity at r = a.
Proof. Without loss of generality we can assume a = 1. If M ≥ −µ * (2) the result follows from [7, Theorem B'], hence we can assume −µ * (1) < M < −µ * (2). We put m = |M | > 0 and If u satisfies (1.13) and blows-up at r = 1, then v = ln u satisfies Up to changing b, we can assuming that u(b) ≥ 1, it follows that v is bounded from below on (1, b) by the solution of (i) We first observe that t and there exists a sequence {t n } converging to 0 such that y(t n ) → ∞. Furthermore z(t) = e Kt y(t) satisfies Therefore y(t) ≥ c 3 t − p+1 p−1 . Using the equation (2.1) we first observe that x t remains negative in a right neighbourhood of 0 otherwise there would exists a sequence {t n } decreasing to 0 where x t (t n ) = 0 and x tt (t n ) ≥ 0 yielding Similarly y t remains negative on some interval (0, τ 1 ) otherwise there would exists a sequence {t n }decreasing to 0 such that y t (t n ) = 0 and y tt (t n ) ≥ 0. Since and x t ≤ 0 we derive a contradiction. Therefore y(t) → ∞ as t → 0, y t ≤ 0 and which implies near t = 0. Using again (1.17) and the monotonicity of x(t) and y(t), p−1 t near t = 0.
(ii) Next we set κ(τ ) = (r − 1) (4.23) and κ and κ τ remain bounded from above and from below on (−∞, a]. Therefore the limit set Σ of the corresponding trajectory at −∞ is not empty and it is included in the limit setΣ at −∞, of some trajectory of a nonnegative function satisfying the autonomous system Which is precisely equation (2.1) in dimension 1. By the Poincaré-Bendixon theorem,Σ either contains an equilibrium or it is a limit cycle. If the limit set contains an equilibrium, say W , it is positive and satisfies Since M > −µ * (1) the only nonnegative root is zero, which yields a contradiction. If the limit set is a cycle γ, it is a subset of Q. This imply that there would exist an equilibrium in the region bordered by γ, contradiction. This ends the proof. Proof. Consider any solution such that (x(0), y(0) = (x 0 , y 0 ) ∈ Q and suppose its negative trajectory T − is defined on some maximal interval (θ, 0] with θ ∈ (−∞, 0) and thus unbounded. We first suppose that t → x(t) is not monotone when t → θ. Then there exists a sequence {t n } decreasing to θ such that x t (t n ) = 0 and thus a n = (x(t n ), y(t n )) ∈ L, x tt (t n ) = y t (t n ) < 0 and lim tn→θ x(t n ) = ∞.
Consider now the regular trajectory T reg . If p ≥ N N −2 then, from Lemma 2.5, either T reg converges to P M when t → ∞, or it crosses L at a point (x * , 2 p−1 x * ) with x * > X M . If n is such that x * < x(t n ) we get a contradiction: indeed for t < t n−1 , T − stays in the region bordered from above by L and T reg , so it cannot intersect L at a n . If 1 < p < N N −2 , we infer the same contradiction using Proposition 4.7. Therefore x(t) decreases monotonically to ∞ when t → θ. Since M > −µ(1) we derive a contradiction. Hence inf I (x 0 ,y 0 ) = −∞. By the same reasons as above, t → x(t) is monotone decreasing and x(t) → ∞ when t → −∞, and y(t) > 2 p−1 x(t) → ∞. Moreover there existst < 0 such that y t (t) ≤ 0: indeed , if for some t 0 <t, y t (t 0 ) > 0, then (x(t 0 ), y(t 0 )) ∈ D and necessarily (x(t), y(t) remains in D for t ≤ t 0 because x t (t) < 0 implies that (x(t), y(t) ∈ D ∪ A and the backward trajectory cannot cross the curve C where y t = 0; now this implies that y t (t) > 0 for t ≤ t 0 , a contradiction. Therefore y t (t) ≤ 0 for t ≤t. Then Since x(t) → ∞, we deduce for large |t|, Returning to the system (1.17), we get for large |t| Since p+1 2 > 1, it is straightforward to check by integration that a positive function x satisfying the above differential inequality cannot be defined on a interval unbounded from below, which ends the proof. Proof. Assume −µ * (1) < M < −µ * (1) + and the regular solution u is not a ground state. Then using the notation of Lemma 4.4 there exists r 0 > 0 such that G(r 0 ) = 0 and G (r 0 ) ≥ 0. Hence u 2 r (r 0 ) = 2p p+1 u p+1 r (r 0 ) = 0, thus Put t 0 = ln r 0 , then y 2 (t 0 ) = 2p p+1 x p+1 (t 0 ) and N − 1 ≤ |y(t 0 )| p−1 p+1 (µ * (1) + M ). Equivalently (4.25) The curve {(x, y) : y 2 = 2p p+1 x p+1 } cuts L at a unique S 0 = (x 0 , y 0 ) ∈ Q and y p−1 . If M = −µ * (1) and p ≥ N N −2 (resp. 1 < p < N N −2 ), then X −µ * (1) > x 0 (resp. X 2,−µ * (1) > x 0 ). Indeed this follows from (2.16). In the same way, if 1 < p < N N −2 , then However, the regular trajectory associated to M , T M reg has a unique intersection with L, at a point (x reg (t 1 ), y reg (t 1 ) where x reg (t 1 ) is maximal, and either p ≥ N N −2 and x reg (t 1 ) > X −µ * (1)+ , or 1 < p < N N −2 and x reg (t 0 ) > X 1,−µ * (1)+ . In both case t 1 < t 0 , y reg (t 1 ) = 2 p−1 x reg (t 1 ) and x reg (t 1 ) > p+1 Proof. Let (x reg (t), y reg (t)) be the regular solution issued from (0, 0). By Lemma 2.8 the function V defined in (2.37) is increasing. Since it vanishes at t = −∞ it is positive. If there exists some t 0 such that x reg (t 0 ) = 0 then V(t 0 ) < 0, which is impossible, hence x reg (t) > 0 for all t. Thus T reg is a ground state; it is bounded by Proposition 2.4, cannot converge to (0, 0) since V(t) > V(−∞) = 0. There is no cycle from Lemma 4.3, hence it converges to P M which is a sink.
Remark. The existence of a ground state was already obtained in [19] with the use of the function Z defined in (2.41).

4.3
The case M < 0, N ≥ 3 and N N −2 < p < N +2 In what follows we give an improvement of [22,Theorem A] in which it is shown that in this range of exponent there exists some ω > 0 such that −ω ≤ −M < 0 there exists no ground state. The expression of ω is explicit (and not simple). Hence, by [17,Th 9.3.3] the homoclinic orbit is repelling. Since P M is also repelling, we derive a contradiction because any trajectory issue from B must converge to T reg . Hence T reg intersects the axis {x = 0} for some positive y 1 > 0, there exists no ground state. We denote by O the region of Q delimited by the regular trajectory T reg and the segment (0, y) : 0 < y < y 1 .
It is negatively invariant. The stable trajectory T st = {x st , y st } of (0, 0) satisfies x st (t) = ce −Kt (1+o(1)) and y st (t) = c(N −2)e −Kt (1+o(1)) when t → −∞, thus it remains in O. Because there are no cycle in O, it must converge to P M , hence the corresponding u st is equivalent to U M near r = 0, which ends to proof.
Remark. In the previous theorem the positive solution u satisfying u(r) ∼ U M (r) as r → 0 such that u(r) ∼ cr 2−N (c > 0) as r → ∞ is the stable trajectory T st . It is a heteroclinic orbit connecting (0, 0) to P M . We conjecture that in the case p = N N −2 the non existence of ground state still holds and that there exists a unique solution u such that satisfying u(r) ∼ U M (r) as r → 0 such that u(r) ∼ cr 2−N (ln r) The expression of the result presents some similarity with Theorem 3.7 in the case M > 0. However a new type of difficulty appears: in order to define properly an intersection function expressing the distance between some trajectories as in [4], we need to find some values of the parameter M for which there exists a ground state. and all the trajectories in Q are bounded. It is not the case when M < −µ * (1) even if there exists a ground state by Lemma 4.4 but we can easily proved that there exist large solutions. So we need to prove that for M = −µ * (1) + there exist a ground state and all the trajectories in Q are bounded. This is the object of Proposition 4.10 and Proposition 4.8. Proof. Recall that M = −µ. First we show that if −µ * (1) < M < M , the stable trajectory T st := T M st either has a limit cycle around P M or does not stay in Q. If we assume that it stays in Q, then it is bounded by Corollary 4.9. Since at −∞ it cannot converge to P M which is a sink by Proposition 4.1, it admits a alpha-limit cycle which is a closed orbit around P M . We present first a general existence result of singular solutions.
Proof. By Proposition 4.2 P 1,M is a saddle point of system (1.17). By the remark after this proposition there exist two stable trajectories T 1,j st , j = 1, 2 converging to P 1,M as t → ∞; the trajectory T 1,1 st is locally below the line L (see Proposition 4.7), hence it belongs to the region C for t > t 0 . By Proposition 4.7 either the regular trajectory T reg converges to P 2,M or it crosses the line L beyond P 2,M . Hence T 1,1 st cannot intersects T reg , and is trapped when t decreases in the region C and the curve T reg . Thus it converges to (0, 0) when t → −∞. Because (0, 0) is a source (see Section 3.4.1) with one fast trajectory T reg , which satisfies lim st is a slow one and it satisfies lim Remark. Under the assumptions of Proposition 4.14 the trajectory T 1,4 unst leaves Q since in this region it stays in the sector {(x, y) : H j (x, y) < 0} for j = 1, 2, and this sector contains no stable equilibrium. The result holds also if M = −µ * .  Then there exist positive real numbers µ <μ min ≤μ max <μ min ≤μ max < µ * (1) with the following properties 1) for µ < |M | <μ min there is no radial ground state; 2) for |M | =μ min or |M | =μ max there exist ground states u satisfying u(r) ∼ U 1,M (r) when r → ∞; 3) for |M | >μ max there exist ground state either such that u(r) ∼ U 2,M (r) when r → ∞ or ondulating around U 2,M (r) when r → ∞.
Moreover st which converges to P 1,M at ∞, cannot converge to P 2,M at −∞. Hence it has a limit cycle around P 2,M . Note that T 1,2 st is included in the region bordered by T 1,3 unst and T 1,4 unst . For α > 0 small enough set M α = −µ * (1)+α. By Proposition 4.10 the corresponding regular trajectory T reg is a ground state. Since (0, 0) is a source and (4.18) holds, it cannot converge neither to (0, 0) nor to P 1,M α as in Lemma 4.4. Hence either it converges to P 2,M α at ∞ or it admits a limit cycle around. Next, T 1,3 unst is trapped in the positively invariant region bordered by T 1,1 st which connects (0, 0) to P 1,M , the portion of the curve C below L (hence between P 1,M and P 2,M ) and T reg between (0, 0) and its second intersection with C; therefore, it converges to P 2,M α at ∞ or it admits a limit cycle too. Finally consider the trajectory T 1,2 st which tends to P 1,M α at ∞. It stays in the negatively invariant region {(x, y) : y > 2 p−1 x or x > X 1,M }. If it stays in Q, then the solution is defined on R and the trajectory is bounded from Corollary 4.9. So, either it converges to a fixed point or it has a limit cycle around P 2,M α when t → −∞. This is impossible since it would intersect T reg . Hence for M = M α , T 1,2 st leaves Q in finite time at (x(t), 0) for some x(t) > 0. Then there exist t 0 > t 1 >t such that y t (t 0 ) = 0 and y(t 0 ) is the maximum of y on (t, ∞) and (x(t 1 ), y(t 1 )) ∈ L. Hence x(t) ≤ x(t 1 ) fort ≤ t ≤ t 1 .
Step 2. Next consider any M ∈ (−µ * (1), M ). In any case the trajectories T 1,2 st and T reg are bounded as long as the stay in Q from Corollary 4.9. Then we define (x M reg , y M reg ) as the farthest point on L belonging to the closure T reg of T reg , and (x M st , y M st ) as the farthest point on L belonging to T st .
Let A be the set of M ∈ (−µ * (1), M ) such that there is no ground state, let B 1 be the set of M ∈ (−µ * (1), M ) such that there exists a ground state converging to P 1,M at ∞ and let B 2 be the set of M ∈ (−µ * (1), M ) such that there exists a ground state converging to P 2,M or having a limit cycle at ∞. Then (−µ * (1), M ) = A ∪ B 1 ∪ B 2 . Clearly M ∈ A and M α ∈ B 2 , furthermore if M ∈ A ∪ B 1 ∪ B 2 we have three possibilities.
• Any M ∈ B 2 has the same properties as M α , hence T reg is a ground state which converges to P 2,M or has a limit cycle around P 2,M ; T 1,3 unst either converges to P 2,M or has a limit cycle around P 2,M ; and T 1,2 st intersects L at a last value t 1 . • If M ∈ A, T reg is not a ground state and it leaves Q through the semi-axis {x = 0, y > 0}; T 1,2 st and T 1,3 unst are included in the region of Q bordered by T reg and three configurations are possible: A-(i) either T 1,2 st has a limit cycle around P 2,M and T 1,3 unst leaves Q A-(ii) or T 1,3 unst converges to P 2,M or has a limit cycle around. A-(iii) or T 1,2 st = T 1,3 unst which means this trajectory is homoclinic with respect to P 1,M . Note that M satisfies A-(i). st and B 1 = ∅. More precisely we can define µ <μ min <μ max < µ * (1) such thatμ min ∈ B 1 ,μ max ∈ B 1 . If |M | <μ min , g(M ) > 0 and there is no ground state. If |M | >μ max , g(M ) < 0 and there exists a ground state u such that r 2 p−1 u(r) → X 2,M or such that r 2 p−1 u(r) is turning around X 2,M as r → ∞. unst starts from P 1,M and converges to P 2,M or has a limit cycle around P 2,M . If −μ max < M < −μ min , h(M ) < 0 and the trajectory T 1,2 st starts from (0, 0) with the slope N − 2 and converges to P 1,M when t → ∞.
Proofs of Theorem C and C'. They are a consequence of Proposition 4.14, Theorem 4.15, Proposition 4.16 and Theorem 4.17.

Remark.
It is an open problem whether the cycles which may exist for some M are unique or not. It is a numerical evidence that it holds if M > 0, but unclear if M < 0.

4.5
The case M < 0, N = 2 and p > 1 A first difficulty in this case comes from the fact that there exist singular solutions u with a logarithmic blow-up. The main difficulty comes from the equality of µ * (2) and µ. Hence −µ is no longer a weak source as in the case N > 2.
3) M = −μ min there exists a ground state u such that u(r) ∼ U 1,M (r) when r → ∞.
Proof. If M = M = −µ * (2) there exists no ground state from Proposition 4.7. By continuity this property is still valid for M = M − for > 0 small enough. As in the proof of Theorem 4.17 with N ≥ 3 we still denote by A the set of M ∈ (−µ * (1), −µ * (2)) such that there is no ground state. We define in a similar way the set B 1 and B 2 . The previous situation is still valid with the only difference that M does not satisfies A-(i) but A-(ii): indeed from [18, Th. 8.2, Lemma 8.7], see Appendix, for < 0 small enough there is no cycle around P 2,M which is a sink by Proposition 4.2. Thus T 1,3 unst converges to P 2 , M when t → ∞ and T 1,2 st converges to 0 when t → −∞, and since there is no ground state it satisfies x(t) ∼ ce Remark. We conjecture that there is no cycle when N = 2. If it is true, then for any −μ < M < −µ * , T 1,2 st converges to (0, 0) as t → −∞. Equivalently there exists a positive solution u of (1.13) with a logarithmic blow-up at r = 0 and such that u(r) ∼ U 1,M (r) as r → ∞. Hence there exist also a positive solution u of (1.13) such that u(r) ∼ U 1,M (r) as r → 0 and u(r) ∼ U 2,M (r) as r → ∞.  Proof. The existence when M < −µ * (1) is proved in Lemma 4.4 but the proof therein is not valid when M = −µ * (1) in which case a second beautiful construction due to Chipot and Weissler [14] applies: if M ≤ −µ * (1) there exist singular solutions U 1,M and U 2,M . If T reg is not a ground state, the corresponding solution u vanishes at r = r 0 > 0. Hence there exists a translation of u, say r → u(r − c) which is tangent to U 1,M which is impossible. If M < −µ * (1) estimate (4.18) implies (4.20) which in turn implies that r 2 p−1 u(r) cannot converge to X 1,M . Notice that there exists no cycle in the phase plane (x, y) otherwise the corresponding solution u would be singular and ondulating hence a translation of it say x → u(x + c) which is now singular at x = −c and defined for x > −c could be made tangent somewhere to U 1,M (or U 2,M ) which is impossible. Therefore r 2 p−1 u(r) converge to X 2,M as r → ∞.
In order to prove that there exists a heteroclinic connecting P 1,M to P 2,M and since T reg converges to P 1,M , there exists a smallest τ such that x reg (τ ) = X 1,M and the vector field H is directed to the right on the segment J = {(x, y) : x = X 1,M , y reg (τ ) ≤ y ≤ Y 1,M }, we have three possibilities: At end we change again time and put in order to see that − → η = (η 1 , η 2 ) verifies  Therefore, the discriminant β 2 2 − 4β 1 of the polynomial P(η 1 ) = β 1 + β 2 η 1 + η 2 1 is positive for α 1 > 0.
By [18,Lemma 8.7] there is no cycles in the region of the plane (β 1 , β 2 ) located in the second quadrant, which is our case since α 1 > 0. Furthermore, the equilibrium (β 1 , β 2 ) = (0, 0), which has a double zero eigenvalue has one stable trajectory converging when t → ∞ and one unstable trajectory converging when t → −∞. Hence it is a saddle point.